If O is the centre of the circle and BCD is tangent to it at C. Prove that angleBAC +angleACD =90 DEGREE (ulike model test paper no. 6 QUESTION NO. 14)
To prove that angle BAC + angle ACD = 90 degrees, we need to use the properties of tangents, circles, and angles. Here's how you can approach the proof.
Step 1: Draw the given figure.
Start by drawing a circle with center O. Within the circle, draw a line segment BCD such that point C is on the circumference, and the line segment is tangent to the circle at point C.
Step 2: Mark the relevant points and angles.
Label the intersection of line segments BC and the circle as point A. Also, label the center of the circle as O.
Step 3: Identify the properties to use.
In this case, we need to consider the following properties:
- The angles formed between a tangent and a radius of a circle are right angles.
- The angle formed by a tangent and a chord in a circle is equal to half the intercepted arc.
Step 4: Proving the statement.
To prove that angle BAC + angle ACD = 90 degrees, we'll break it down into two parts.
Part 1: Prove that angle BAC = 90 degrees.
Since triangle BOC is a right angle triangle (formed by the radius and tangent), we know that angle BOC = 90 degrees.
Since angle BOC is an inscribed angle (formed by points B, O, and C on the circle), the intercepted arc BC will also be 90 degrees.
Since angle BAC is equal to half of the intercepted arc BC, we can conclude that angle BAC = 90 degrees.
Part 2: Prove that angle ACD = 90 degrees.
Since BD is a line segment tangent to the circle at point C, and angle BCD is 90 degrees (formed between the tangent and radius), we can conclude that angle BCD = angle ACD = 90 degrees.
Step 5: Summing up the angles.
By adding angle BAC and angle ACD, we get:
angle BAC + angle ACD = 90 degrees + 90 degrees = 180 degrees.
Therefore, angle BAC + angle ACD = 180 degrees, which implies the given statement: angle BAC + angle ACD = 90 degrees.